Let , , , , , and be positive integers. We consider that , and have no common prime factors and , and are the edge lengths, , and are the diagonal lengths and is the space diagonal length. If a perfect cuboid exists, the following equations hold.
is odd and and are even because , and are not all even and if two edges are odd, then the diagonal is not a square number. And and are odd and is even. Let , , and be positive inetgers and hods. We consider that and are relatively prime and the odd/even number of and are different. By the form of Pythagorean triple, the following equations hold.
is even and the others are all odd. We consider that , and are mutually prime to each other since if two of the diagonal lengths have a common prime factor, then the length of the space diagonal and the length of the other diagonal will also have the prime factor, three edge lengths will have it too and this contradicts the condition that , and have no common prime factors. Let be an odd positive integer. Since has factors , and ,
holds. Let and be odd positive integers and be an even positive integer. By this equation, the following equations hold.
By these equations,
holds. Let and be odd positive integers and be an even positive integer. By this equation, the following equations hold.
By these equations,
holds. Comparing the equation (3) with the equation (2), we find that the following equations hold since the equations (2) and (3) must hold for any , and satisfying the equation (1).
It becomes a contradiction since and are odd. From the above, there are no perfect cuboids. (Q.E.D.)